. . . . . .

Section 2.5The Chain Rule

V63.0121, Calculus I

February 19, 2009

Announcements

I Midterm is March 4/5 (75 min., in class, covers 1.1–2.4)I ALEKS is due February 27, 11:59pm

. . . . . .

Outline

Compositions

HeuristicsAnalogyThe Linear Case

The chain rule

Examples

Related rates of change

. . . . . .

CompositionsSee Section 1.2 for review

DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

.

.g .f.x .g(x) .f(g(x))

.f ◦ g

Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.

. . . . . .

CompositionsSee Section 1.2 for review

DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g

.f

.x .g(x)

.f(g(x)).f ◦ g

Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.

. . . . . .

CompositionsSee Section 1.2 for review

DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x)

.f(g(x)).f ◦ g

Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.

. . . . . .

CompositionsSee Section 1.2 for review

DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.

. . . . . .

CompositionsSee Section 1.2 for review

DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.

. . . . . .

CompositionsSee Section 1.2 for review

DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.

. . . . . .

Outline

Compositions

HeuristicsAnalogyThe Linear Case

The chain rule

Examples

Related rates of change

. . . . . .

Analogy

Think about riding a bike. Togo faster you can either:

I pedal fasterI change gears

.

.Image credit: SpringSun

The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):

φ(θ) =Rθ

r

And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.

. . . . . .

Analogy

Think about riding a bike. Togo faster you can either:

I pedal faster

I change gears

.

.Image credit: SpringSun

The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):

φ(θ) =Rθ

r

And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.

. . . . . .

Analogy

Think about riding a bike. Togo faster you can either:

I pedal fasterI change gears

.

.Image credit: SpringSun

The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):

φ(θ) =Rθ

r

And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.

. . . . . .

Analogy

Think about riding a bike. Togo faster you can either:

I pedal fasterI change gears

.

.Image credit: SpringSun

The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):

φ(θ) =Rθ

r

And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.

. . . . . .

The Linear Case

QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?

Answer

I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the

two functions.

The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.

. . . . . .

The Linear Case

QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?

Answer

I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)

I The composition is also linearI The slope of the composition is the product of the slopes of the

two functions.

The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.

. . . . . .

The Linear Case

QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?

Answer

I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linear

I The slope of the composition is the product of the slopes of thetwo functions.

The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.

. . . . . .

The Linear Case

QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?

Answer

I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the

two functions.

The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.

. . . . . .

The Linear Case

QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?

Answer

I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the

two functions.

The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.

. . . . . .

The Linear Case

QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?

Answer

I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the

two functions.

The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.

. . . . . .

Outline

Compositions

HeuristicsAnalogyThe Linear Case

The chain rule

Examples

Related rates of change

. . . . . .

Theorem of the day: The chain rule

TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and

(f ◦ g)′(x) = f′(g(x))g′(x)

In Leibnizian notation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Observations

I Succinctly, the derivative of acomposition is the product ofthe derivatives

I The only complication iswhere these derivatives areevaluated: at the same pointthe functions are

I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions

.

.Image credit: ooOJasonOoo

. . . . . .

Theorem of the day: The chain rule

TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and

(f ◦ g)′(x) = f′(g(x))g′(x)

In Leibnizian notation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Observations

I Succinctly, the derivative of acomposition is the product ofthe derivatives

I The only complication iswhere these derivatives areevaluated: at the same pointthe functions are

I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions

.

.Image credit: ooOJasonOoo

. . . . . .

CompositionsSee Section 1.2 for review

DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.

. . . . . .

Observations

I Succinctly, the derivative of acomposition is the product ofthe derivatives

I The only complication iswhere these derivatives areevaluated: at the same pointthe functions are

I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions .

.Image credit: ooOJasonOoo

. . . . . .

Theorem of the day: The chain rule

TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and

(f ◦ g)′(x) = f′(g(x))g′(x)

In Leibnizian notation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Theorem of the day: The chain rule

TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and

(f ◦ g)′(x) = f′(g(x))g′(x)

In Leibnizian notation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Outline

Compositions

HeuristicsAnalogyThe Linear Case

The chain rule

Examples

Related rates of change

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, write h as f ◦ g. Let f(u) =

√u and g(x) = 3x2 + 1. Then

f′(u) = 12u−1/2, and g′(x) = 6x. So

h′(x) = 12u−1/2(6x) = 1

2(3x2 + 1)−1/2(6x) =3x√

3x2 + 1

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, write h as f ◦ g.

Let f(u) =√

u and g(x) = 3x2 + 1. Thenf′(u) = 1

2u−1/2, and g′(x) = 6x. So

h′(x) = 12u−1/2(6x) = 1

2(3x2 + 1)−1/2(6x) =3x√

3x2 + 1

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, write h as f ◦ g. Let f(u) =

√u and g(x) = 3x2 + 1.

Thenf′(u) = 1

2u−1/2, and g′(x) = 6x. So

h′(x) = 12u−1/2(6x) = 1

2(3x2 + 1)−1/2(6x) =3x√

3x2 + 1

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, write h as f ◦ g. Let f(u) =

√u and g(x) = 3x2 + 1. Then

f′(u) = 12u−1/2, and g′(x) = 6x. So

h′(x) = 12u−1/2(6x)

= 12(3x2 + 1)−1/2(6x) =

3x√3x2 + 1

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, write h as f ◦ g. Let f(u) =

√u and g(x) = 3x2 + 1. Then

f′(u) = 12u−1/2, and g′(x) = 6x. So

h′(x) = 12u−1/2(6x) = 1

2(3x2 + 1)−1/2(6x) =3x√

3x2 + 1

. . . . . .

Does order matter?

Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solution

I For the first, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I For the second, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · sin x

. . . . . .

Does order matter?

Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solution

I For the first, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I For the second, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · sin x

. . . . . .

Does order matter?

Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solution

I For the first, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I For the second, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · sin x

. . . . . .

Order matters!

Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solution

I For the first, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I For the second, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · sin x

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2

. Find f′(x).

Solution

ddx

(3√

x5 − 2 + 8)2

= 2(

3√

x5 − 2 + 8) d

dx

(3√

x5 − 2 + 8)

= 2(

3√

x5 − 2 + 8) d

dx3√

x5 − 2

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3 d

dx(x5 − 2)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2

. Find f′(x).

Solution

ddx

(3√

x5 − 2 + 8)2

= 2(

3√

x5 − 2 + 8) d

dx

(3√

x5 − 2 + 8)

= 2(

3√

x5 − 2 + 8) d

dx3√

x5 − 2

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3 d

dx(x5 − 2)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2

. Find f′(x).

Solution

ddx

(3√

x5 − 2 + 8)2

= 2(

3√

x5 − 2 + 8) d

dx

(3√

x5 − 2 + 8)

= 2(

3√

x5 − 2 + 8) d

dx3√

x5 − 2

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3 d

dx(x5 − 2)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2

. Find f′(x).

Solution

ddx

(3√

x5 − 2 + 8)2

= 2(

3√

x5 − 2 + 8) d

dx

(3√

x5 − 2 + 8)

= 2(

3√

x5 − 2 + 8) d

dx3√

x5 − 2

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3 d

dx(x5 − 2)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2

. Find f′(x).

Solution

ddx

(3√

x5 − 2 + 8)2

= 2(

3√

x5 − 2 + 8) d

dx

(3√

x5 − 2 + 8)

= 2(

3√

x5 − 2 + 8) d

dx3√

x5 − 2

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3 d

dx(x5 − 2)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2

. Find f′(x).

Solution

ddx

(3√

x5 − 2 + 8)2

= 2(

3√

x5 − 2 + 8) d

dx

(3√

x5 − 2 + 8)

= 2(

3√

x5 − 2 + 8) d

dx3√

x5 − 2

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3 d

dx(x5 − 2)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

A metaphor

Think about peeling an onion:

f(x) =

(3√

x5︸︷︷︸�5

−2

︸ ︷︷ ︸3√�

+8

︸ ︷︷ ︸�+8

)2

︸ ︷︷ ︸�2

.

.Image credit: photobunny

f′(x) = 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

. . . . . .

Combining techniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)

SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:

ddx

((x3 + 1)10 · sin(4x2 − 7)

)=

(ddx

(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·

(ddx

sin(4x2 − 7)

)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)

. . . . . .

Combining techniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:

ddx

((x3 + 1)10 · sin(4x2 − 7)

)=

(ddx

(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·

(ddx

sin(4x2 − 7)

)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)

. . . . . .

Combining techniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:

ddx

((x3 + 1)10 · sin(4x2 − 7)

)=

(ddx

(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·

(ddx

sin(4x2 − 7)

)

= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)

. . . . . .

Combining techniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:

ddx

((x3 + 1)10 · sin(4x2 − 7)

)=

(ddx

(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·

(ddx

sin(4x2 − 7)

)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)

. . . . . .

Outline

Compositions

HeuristicsAnalogyThe Linear Case

The chain rule

Examples

Related rates of change

. . . . . .

Related rates of change

QuestionThe area of a circle, A = πr2,changes as its radius changes. Ifthe radius changes withrespect to time, the change inarea with respect to time is

A.dAdr

= 2πr

B.dAdt

= 2πr +drdt

C.dAdt

= 2πrdrdt

D. not enough information

.

.Image credit: Jim Frazier

. . . . . .

Related rates of change

QuestionThe area of a circle, A = πr2,changes as its radius changes. Ifthe radius changes withrespect to time, the change inarea with respect to time is

A.dAdr

= 2πr

B.dAdt

= 2πr +drdt

C.dAdt

= 2πrdrdt

D. not enough information

.

.Image credit: Jim Frazier